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For $t>0$, write $\Phi$ for the standard normal distribution function. Since the Brownian running maximum is nonnegative, $a\le 0$ gives simply $\mathbb{P}(B_t\le b)=\Phi (b/\sqrt{t})$.

For $a>0$ and $b<a$, reflect the Brownian motion after its first hit of $a$. This is a stopping time, and the Strong Markov property together with symmetry of Brownian increments shows that reflection preserves its law. On paths that hit $a$, it sends the endpoint $B_t$ to $2a-B_t$. Thus the event with endpoint at most $b$ maps to endpoints at least $2a-b$, all of which necessarily hit $a$. This provesThe same reflection gives $\mathbb{P}(M_t\ge a)=2(1-\Phi (a/\sqrt{t}))$. If $b\ge a$, every path with $B_t>b$ has hit $a$, so subtracting that endpoint tail gives the complete answer:The expressions agree at $b=a$. This is the joint distribution of Brownian motion and its running maximum. In particular $M_t$ has the distribution of $|B_t|$ and has no atoms for $t>0$. If $t=0$, the pair is $(0,0)$ deterministically, so the requested probability is $1_{\{a\le 0,b\ge 0\}}$.

Fix $0<t<1$. The event $\{\tau \le t\}$ means that the future path through time one never exceeds the maximum already attained by time $t$. Given ${\mathcal{F}}_t$, future increments form an independent Brownian motion $W$, soHere the Brownian running maximum distribution is the one derived above. On $\{B_t<0\}$, an event of probability $1/2$, the finite quantity $M_t-B_t$ is strictly positive, so this conditional probability lies strictly between zero and one.

If $\tau$ were a stopping time, $\{\tau \le t\}$ would be ${\mathcal{F}}_t$-measurable, and the displayed conditional expectation of its indicator would equal that indicator, taking only zero and one. The contradiction proves $\tau$ is not an $({\mathcal{F}}_t)$-stopping time. Thus the Strong Markov property cannot be invoked at $\tau$ to assert a Brownian shifted process; the direct argument here does not assume that property at this random time.

Put $Y_t=B_{\tau +t}-B_{\tau }$. Because $B_{\tau }$ is the maximum on $[0,1]$,The preceding part makes this a strictly positive random interval with probability one. A standard Brownian motion cannot have this path property. For each deterministic $h>0$, the Brownian reflection principle gives $\mathbb{P}({\sup }_{0\le t\le h}{B}_t\le 0)=0$. Taking the countable union over $h=1/k$ shows that the probability of staying nonpositive on any initial interval of positive length is zero.

Therefore $Y$ is not a Brownian motion. This is the obstruction for Brownian motion shifted at its finite-horizon maximum. It concerns its path law, irrespective of any proposed filtration for the shifted process.

Continuity ensures that the maximum on $[0,1]$ is attained, so its first attainment time $\tau$ lies in $[0,1]$. To exclude the endpoint, use Brownian time reversal on a finite interval:This is a standard Brownian motion on this interval. Indeed, its increments are increments of $B$ on disjoint intervals taken in reverse order, hence are independent centered normal variables with the required variances, and its paths are continuous.

If $B_1$ is the maximum, then $R_s\ge 0$ for every $s$. But by the Brownian reflection principle and symmetry,the running maximum of $-R$ has the distribution of $|R_1|$, whose probability of being zero is zero. In particular the event $\tau =1$ has probability zero. Thus $\tau <1$ with probability one. Uniqueness of the maximizing time is not needed for this proof.

Use the Kolmogorov continuity theorem in its one-parameter form: if on a compact interval a process satisfies $\mathbb{E}|X_t-X_s{|}^p\le C|t-s{|}^{1+\beta }$ for some $p,\beta >0$, it has a modification whose paths are Hölder continuous of every order $\gamma <\beta /p$ on that interval.

For Brownian motion, normal increments give, for every $p>0$,Every such normal moment is finite. Taking $p>2$ yields $\beta =p/2-1$, hence any order below $1/2-1/p$. For a prescribed $0<\alpha <1/2$, choose $p$ with $1/2-1/p>\alpha$.

The continuous modification and the given continuous Brownian motion agree at all rational times on one almost sure event; continuity makes them agree everywhere on the interval. To obtain all exponents and all compact intervals simultaneously, apply the theorem to integer intervals $[0,N]$ and a countable sequence of positive exponents increasing to $1/2$, then intersect these almost sure events. A bound at exponent $\gamma >\alpha$ implies a bound at $\alpha$ on a compact interval. Consequently the Brownian Hölder regularity conclusion iswith finite random constants on one common event of probability one. This uses the usual positive-exponent meaning of Hölder continuity.

A finite right derivative at zero would make the difference quotients $B_{1/n}/(1/n)$ eventually bounded. Therefore that event is contained inFor fixed $K,N$, its probability is at most $\mathbb{P}(|B_{1/n}|\le K/n)$ for each $n\ge N$. By the normal distribution of the Brownian increment,Thus every event in this countable union has probability zero. With probability one, Brownian motion has no finite right derivative at zero, which is the appropriate derivative for its time domain. No independence of the quotients is assumed or needed. This proves the required endpoint case of nowhere differentiability of Brownian motion.

Use the Donsker invariance principle: for independent and identically distributed random variables of mean zero and variance one, the linearly interpolated processes $W_n$ with $W_n(k/n)=S_k/\sqrt{n}$ converge in distribution to standard Brownian motion in $C[0,1]$ with the uniform topology. No moment beyond the finite second moment is required.

The maximum functional $F(f)={\max }_{0\le t\le 1}f(t)$ is continuous, since $|F(f)-F(g)|\le \Vert f-g{\Vert }_{\infty }$. A linear function on each interpolation interval has its maximum at an endpoint, so $F(W_n)=M_n/\sqrt{n}$. The continuous mapping theorem therefore givesBy the Brownian reflection principle, the limiting random variable has the distribution of $|Z|$, $Z\sim N(0,1)$. Its distribution function is continuous and has no atom at any $x\ge 0$, so convergence in distribution permits passage to these tail probabilities. HenceAt $x=0$ the left side is exactly one for every $n$, because $S_0=0$ is included in the maximum, and the right side is also one.

Use $V=⟨X⟩$. Applying the Itô formula and the Itô product rule givesThe finite-variation terms cancel in the prescribed combination, leavingThus $Y$ is initially a continuous local martingale. It is a true martingale, not merely local. On each fixed horizon $T$,Part (a) with $p=4$, the assumed bracket moments and Cauchy-Schwarz inequality make this bound integrable. The integrable-supremum martingale criterion now applies, by localization and dominated conditional expectations. The same reasoning makes $X^2-V$ a martingale, so $\mathbb{E}{V}_t=\mathbb{E}{X}_t^2=t$.

The zero covariance now means $\mathbb{E}(X_t^2{V}_t)=t^2$. Since $Y_0=0$ and $\mathbb{E}{Y}_t=0$,This is the fourth-moment deficit and bracket variance identity. If equality holds for every $t$, then $V_t=t$ almost surely for each $t$. Take a single probability-one event for all rational times and use continuity to obtain $V_t=t$ simultaneously for all times. The Lévy characterization of Brownian motion then proves $X$ is Brownian motion in its given filtration. A proof of that characterization by conditional characteristic functions is included in Question 2(a).

Let $V_t=⟨M{⟩}_t$, and first suppose $V_{\infty }=\infty$ almost surely in addition to strict increase. The Dambis-Dubins-Schwarz theorem says thatdefine a Brownian motion $B$ in the time-changed filtration and giveEach $T_u$ is a stopping time and is finite. Continuity and strict increase of $V$ make $u\mapsto T_u$ continuous, $V_{T_u}=u$ and $T_{V_t}=t$.

Here are the martingale details behind this inverse-clock proof of the Dambis-Dubins-Schwarz theorem. Stopping a continuous local martingale when its bracket reaches $n$ makes it an L2-bounded continuous martingale. This follows from the stopped Itô isometry or from the $p=2$ estimate proved in Question 1(a), applied after localization. In particular it is uniformly integrable, and optional sampling is valid even at an unbounded stopping time by taking limits. Applying this to $M$ stopped at $T_n$ shows that $(M_{T_{u\wedge n}}{)}_{u\ge 0}$ is a martingale. Thus $B$ is a continuous local martingale. Time-changing $M^2-V$ in the same way shows $B_u^2-u$ is a local martingale, so $⟨B{⟩}_u=u$.

For completeness, the Lévy characterization of Brownian motion follows directly from the Itô formula. If a continuous local martingale $N$, starting at zero, has bracket $u$, thenis a complex local martingale. Its modulus is bounded on each deterministic finite horizon, so it is a true martingale there. ConsequentlyConditional characteristic functions give Gaussian increments independent of the past. Iterating this identity gives independent increments, and continuity completes the Brownian characterization. The identical vector argument proves the Lévy characterization of multidimensional Brownian motion when the bracket matrix is $uI$.

The printed strict-increase hypothesis does not imply $V_{\infty }=\infty$. For example, $M_t={\int }_0^t{e}^{-s}d{W}_s$ has strictly increasing bracket $(1-e^{-2t})/2$. To state the theorem under exactly the printed hypothesis, allow an independent enlargement of the probability space if the terminal clock $L=V_{\infty }$ can be finite.

On $\{L<\infty \}$ the martingale $M$ has a finite terminal limit. Indeed, stopping at each bracket level $n$ gives an L2-bounded continuous martingale which converges; on $\{V_{\infty }<n\}$ the stopped process is the original one. This proves the finite-bracket convergence lemma. Set $T_u=\infty$ when $u\ge L$ and continue $N_u=M_{T_u}$ by that terminal limit. The optional-sampling argument just given makes $N$ a continuous local martingale with bracket $u\wedge L$. Moreover $L$ is a stopping time in ${\mathcal{G}}_u={\mathcal{F}}_{T_u}$.

On a product extension add an independent Brownian motion $\beta$ in clock time, and putThe two summands have zero quadratic covariation, and their brackets are $u\wedge L$ and $(u-L{)}^{+}$. Thus $⟨\tilde{B}{⟩}_u=u$; the proved characterization makes $\tilde{B}$ Brownian. Since $V_t<L$ at every finite $t$ when $L$ is finite, $M_t={\tilde{B}}_{V_t}$ still holds. This is the finite-lifetime extension of the Dambis-Dubins-Schwarz theorem. An infinite clock gives Brownian motion on the original space; a finite clock may require the independent extension.

Define the radial martingale and its clock byThe Itô formula gives $R_t^2=2H_t+2t$. Independence of the coordinate Brownian motions makes their cross variation zero, soThe last identity is orthogonality of the radial martingale and planar Brownian area.

The clock is adapted and continuous, and it is strictly increasing almost surely. Otherwise the two coordinate paths would both vanish throughout a nontrivial interval. Such an interval contains a rational subinterval, while a Brownian increment over each fixed rational subinterval is a nondegenerate Gaussian and cannot be zero with positive probability.

Also $A(\infty )=\infty$ almost surely. If it were finite, the finite-bracket convergence lemma would make $H_t$ converge to a finite limit. Then $R_t^2=2t+O(1)$ on that event, forcing ${\int }_0^{\infty }{R}_t^{2}dt=\infty$, a contradiction. Thus no finite-lifetime extension is needed here.

Use the same inverse clock $T_u=\inf \{t:A(t)>u\}$ for both martingales, and set $W_u=H_{T_u}$, $B_u=Z_{T_u}$. Their bracket matrix isThe vector characterization proved in part (a) makes $(W,B)$ a two-dimensional Brownian motion; in particular its two coordinate processes are independent. Reversing the common clock givesThis is a common-clock Brownian representation of radius and area. Independence follows from the joint time change and identity bracket matrix; no independence of either Brownian motion from $A$ is asserted.

Let $S$ denote this maximum and put $\tau =\tau (-b)$. Brownian motion reaches $-b$ almost surely: the Brownian reflection principle gives crossing probability $2\mathbb{P}(W_t<-b)\to 1$. Continuity gives $W_{\tau }=-b$, with the strict-crossing infimum interpreted as in part (b). The processis a continuous nonnegative local martingale starting at one and tending to zero. Its maximum is $1+S/b$. Apply part (b) at $a=1+x/b$, for $x>0$:Differentiating gives the maximum before a lower Brownian barrier densityThe tail tends to one as $x\downarrow 0$, so there is no atom at zero. The density integrates to one.

Use the same Brownian motion for both solutions: this is a synchronous coupling. For $D_t=X_t-Y_t$, the Itô formula givesOn any fixed finite horizon the stochastic integral has mean zero. Indeed, boundedness of $\sigma$ and the assumed second-moment bounds make the expectation of its squared integrand integrable in time. With $h(t)=\mathbb{E}{D}_t^2$, the contraction condition yields, for every $s\le t$,The allowed Gronwall inequality gives the mean-square contraction of synchronously coupled diffusionsThe identical estimate can also be obtained by localizing the nonnegative local supermartingale $e^{kt}{D}_t^2$ and using Fatou's lemma, a formulation useful when the drift has linear growth.

There is a genuine compatibility issue in the printed global assumptions. If $|b|\le B$, then for $d=|x-y|$,It cannot be at most $-k{d}^2$ for all $d>4B/k$. Thus a globally bounded drift cannot satisfy the stated strict contraction on all of the real line. The stochastic estimate above is the requested conditional calculation. For a non-vacuous application, global boundedness of the drift must be relaxed while retaining suitable existence and moment hypotheses; it is not silently changed here.

Enlarge the space if needed to choose $Y_0$ with density $p$, independent of the driving Brownian motion, and solve the same equation with the same driver as $X_0=x$. The second moment of $p$ makes the initial difference square integrable. By parts (a) and (b),while part (c) givesWe must not assume that the smooth bounded $f$ is globally Lipschitz. The hypotheses give $C_Y={\sup }_t\mathbb{E}{Y}_t^2<\infty$, hence uniform tightness of $(Y_t)$. For any $R$ and $0<\delta \le 1$, define the modulus of continuity of $f$ on $[-R-1,R+1]$ by ${\omega }_R(\delta )$. Splitting according to $|Y_t|\le R$ and $|X_t-Y_t|\le \delta$, and applying Markov inequality, givesFirst choose $R$ large, then $\delta$ small using uniform continuity on that compact interval, and finally let $t\to \infty$. The right side can be made arbitrarily small. Together with part (a), this provesThis is convergence by synchronous coupling for bounded continuous test functions. It only needs the constant expectation from part (b) and uniform second moments; it does not add an unstated global bound on $f^{\prime }$ or require a separate stationarity theorem. As explained in part (c), the literal bounded-drift plus strict-contraction assumptions have no global example; the calculation records the intended consequence under compatible dissipative-drift hypotheses as well.

A compact H-hull is a bounded set $K\subset \mathbb{H}$ that is closed relative to the complex upper half-plane and whose complement $D=\mathbb{H}\setminus K$ is a simply connected domain. Its Euclidean closure is compact in $\overline{\mathbb{H}}$; “compact” here does not require separation from the real axis. One may equivalently describe the closed hull in $\overline{\mathbb{H}}$, with its real-boundary convention understood. Its mapping-out function is the unique conformal map to $\mathbb{H}$ with hydrodynamic normalization at infinity.

Use the capacity convention $\mathrm{hcap}(K_t)=2t$. A chordal Schramm–Loewner evolution from $0$ to infinity is the hull family of the Chordal Loewner equationwhere $W$ is a real standard Brownian motion and $\kappa \ge 0$. The points with interior-point swallowing time for a Loewner chain at most $t$ constitute $K_t$. For each $t$, $g_t$ maps $\mathbb{H}\setminus K_t$ conformally onto $\mathbb{H}$ and has expansion $g_t(z)=z+2t/z+O(z^{-2})$. The case $\kappa =0$ is the deterministic vertical-slit chain.

First identify the relevant transformed Loewner driving functions directly. For $\lambda >0$, defineDifferentiation gives ${\partial }_t{\tilde{g}}_t=2/({\tilde{g}}_t-{\lambda }^{-1}{U}_{{\lambda }^{2}t})$. Its initial value is $z$, its domain is $\mathbb{H}\setminus {\lambda }^{-1}{K}_{{\lambda }^{2}t}$, and its expansion is $z+2t/z+O(z^{-2})$. Thus its driving function is ${\lambda }^{-1}{U}_{{\lambda }^{2}t}$, by uniqueness of the differential equation and of the hydrodynamically normalized map. Brownian scaling provesThis is capacity-parametrized scale invariance of a Loewner chain.

For a fixed time $t$, map the future remaining domains by $z\mapsto g_t(z)-U_t$. The corresponding hulls ${\hat{K}}_s$ are specified without any boundary-image ambiguity byTheir mapping-out functions areAt $s=0$ this is the identity. Differentiating at a point in its domain yieldsThe Laurent series is $z+2s/z+O(z^{-2})$. Hence the transformed Loewner driver is exactly $U_{t+s}-U_t$, with the original capacity clock unchanged. This proves the composition rule for chordal Loewner driving functions, rather than assuming an identification of transforms.

The Brownian motion increments after $t$ are independent of its past and have the original law. Since past hulls are measurable functions of the past Loewner driver, conditionally on the past, the centered mapped future is an independent copy of the original chain. This is the domain Markov property of a chordal Loewner chain. At an almost surely finite stopping time it follows likewise from the Strong Markov property.

For the converse, a precise class is essential: take capacity-parametrized Loewner chains generated by a continuous real Loewner driving function $U$, with $U_0=0$, and require the two displayed hull properties, including independence from the entire hull past in the domain Markov property. Within this class they characterize ${\mathrm{SLE}}_{\kappa }$, for some $\kappa \ge 0$.

To see why the hull assertions determine the Loewner driver assertions, the hull domains determine their normalized maps uniquely. For any surviving point,The left time derivative suffices for $t>0$, since $U$ is continuous. A sufficiently high point survives up to any given finite time, so points $in$, $n\in \mathbb{N}$, suffice to reconstruct the Loewner driver locally and measurably from the hull past. Thus the driving-function reconstruction for a chordal Loewner chain identifies the two past filtrations and makes the preceding scale and composition computations reversible.

The domain Markov property therefore gives stationary independent increments for $U$. Its continuity makes it a continuous Lévy process. Its classification gives $U_t=bt+\sqrt{\kappa }{W}_t$: in the Lévy–Khintchine formula continuity removes the jump measure, leaving characteristic function $\exp (t(ibu-\kappa u^2/2))$. Driver scaling would turn the drift $b$ into $\lambda b$, so invariance for every $\lambda$ forces $b=0$. The diffusion coefficient is unchanged. This proves the scale-and-domain-Markov characterization of SLE with its regularity and parametrization hypotheses stated explicitly.